I am trying to understand the connections between holomorphic vector (line) bundles, sections and sheaves in the simplest terms possible and I want to check that my interpretations are correct. I am somewhat familiar with bundles but certainly lack background in sheaves.
Goal/question: correct everything that might be wrong in the following interpretation, any nitpick will be welcome. More concretely: verify that for the sheaf morphism given below, the kernel $\ker \phi$ vanishes.
Consider $X=\mathbb{P}^1_\mathbb{C}$ the projective line, and two line bundles over it: $\mathcal{O}$, the trivial line bundle, and $\mathcal{O}(1)$, the hyperplane bundle. I wanted to make a connection between global sections $\phi\in H^0(\mathbb{P}^1,\mathcal{O}(1))$ and sheaf morphisms $$ \mathcal{O}\overset{\phi}{\rightarrow}\mathcal{O}(1). $$ Let us take a particularly simple one: $\phi\equiv z_0$, a homogeneous polynomial of degree $1$ in $(z_0,z_1)$.
1: Interpret line bundles as sheaves: for any line bundle $L$ of the above and any open set $U\subset\mathbb{P}^1$ interpret define $\Gamma(U,L)$ to be the $\mathcal{O}(U)$-module of local sections of the line bundle $L$ over $U$. This defines a presheaf that turns out to be a sheaf.
2: Interpret $\phi$ as a sheaf morphism: for any small open set (i.e. $U$ contained in a trivializing open set of both bundles) I can map local sections of $\mathcal{O}$ to local sections of $\mathcal{O}(1)$ by multiplying by the polynomial $\phi$ in the corresponding homogeneous coordinates. In terms of the trivializations:
Over $U_0=\{(z_0:z_1)|z_0\neq 0\}$, the section $\phi= z_0:U_0\rightarrow \mathcal{O}(1)$ is written in terms of the trivialization $\mathcal{O}(1)(U_0)\simeq U_0\times \mathbb{C}$ as $\phi(\xi_0:\xi_1) \equiv ((\xi_0,\xi_1),1)$. Therefore $\phi$ as a map $\Gamma (U_0,\mathcal{O})\rightarrow \Gamma(U_0,\mathcal{O}(1))$ sends sections $f$ of $\mathcal{O}(U_0)\simeq U_0\times \mathbb{C}$ (which are nothing but holomorphic functions of $\frac{z_1}{z_0}$) to themselves. Clearly an isomorphism of $\mathcal{O}(U_0)$-modules.
Over $U_1=\{(z_0:z_1)|z_1\neq 0\}$, the section $\phi= z_0:U_1\rightarrow \mathcal{O}(1)$ is written in terms of the trivialization $\mathcal{O}(1)(U_1)\simeq U_1\times \mathbb{C}$ as $\phi(\xi_0:\xi_1) \equiv ((\xi_0,\xi_1),\frac{\xi_0}{\xi_1})$. Therefore $\phi$ as a map $\Gamma(U_1,\mathcal{O})\rightarrow \Gamma(U_1,\mathcal{O}(1))$ sends sections $f$ of $\mathcal{O}(U_1)\simeq U_1\times \mathbb{C}$ (holomorphic functions of $\frac{z_0}{z_1}$) to themselves $f(z_0/z_1)\cdot \frac{z_0}{z_1}$. This is not a surjection of sections but is certainly injective: only the zero section is mapped to the zero section.
The problem arises when considering bigger open sets, for example $\mathbb{P}^1$ itself. In that case, I do not have such a simple interpretation of sections as holomorphic function of some homogenized coordinates and it certainly cannot be an isomorphism of $\mathbb{C}$-vector spaces: $$ \mathbb{C}\simeq \Gamma(\mathbb{P}^1,\mathcal{O}) \rightarrow \text{span}_\mathbb{C}\{z_0,z_1\}\simeq \Gamma(\mathbb{P}^1,\mathcal{O}(1)) $$ simply by counting the dimensions!
3: Consider kernel and quotient sheaves: based on the calculations above, I think that as a morphism of sheaves, $\phi$ is injective, meaning that even though it vanishes at a point (0:1), no nonzero local section is mapped to a local zero section. This would imply that the sheaf $\ker \phi = 0$.
Much more difficult would be to construct the cokernel sheaf $\mathcal{O}(1) / \text{im }\phi$. It will certainly not correspond to a line bundle as it is impossible that it is locally free.